Question

Let $y$ be an implicit function of $x$ defined by ${{x}^{2x}}-2{{x}^{x}}\cot y-1=0$ . Then, ${{y}^{'}}(1)$ equals:
(a) $1$
(b) $\log 2$
(c) $-\log 2$
(d) $-1$

Answer 1

Hint:The given problem is related to differentiation of implicit function. Use the following formulae to get the required answer: $\dfrac{d(\cot x)}{dx}=-\cos e{{c}^{2}}x$ and $\dfrac{d({{x}^{x}})}{dx}={{x}^{x}}(1+\log x)$ .

Complete step-by-step answer:
The given expression is ${{x}^{2x}}-2{{x}^{x}}\cot y-1=0$ .
The given expression can be rewritten as ${{({{x}^{x}})}^{2}}-2{{x}^{x}}\cot y-1=0$ .
Clearly , we can see that it is an implicit function . Hence , we can represent it as $f(x,y)\equiv {{({{x}^{x}})}^{2}}-2{{x}^{x}}\cot y-1=0$.
Now , we will differentiate $f(x,y)$ with respect to $x$ .
$f(x,y)$ is an implicit function. So , to differentiate it , we will apply chain rule of differentiation . According to the chain rule of differentiation , if $R(x,y)$ is an implicit function , then the derivative of $R(x,y)$ with respect to $x$ is given by $\dfrac{dR}{dx}=\dfrac{\partial R}{\partial x}+\dfrac{\partial R}{\partial y}\times \dfrac{dy}{dx}$ .
Now , we will differentiate the function with respect to $x$ . So, on differentiating we get ,
$2({{x}^{x}}).\dfrac{d}{dx}({{x}^{x}})-\cot y.\dfrac{\partial }{\partial x}2({{x}^{x}})-(2{{x}^{x}}\dfrac{\partial }{\partial y}(\cot y)\times \dfrac{dy}{dx})=0.........$ equation $(1)$
Now, we have to find the derivative of ${{x}^{x}}$ with respect to $x$ .
Let z = ${{x}^{x}}$ . Since, the function is in exponent form, let’s apply logarithms to convert it into product form. So, log z = log $\left( {{x}^{x}} \right)$ . We know, $\log {{a}^{b}}=b\log a$.
$\Rightarrow \log z=x\log x$ . Now, to differentiate with respect to x, we will use product rule of differentiation, which is given as $\dfrac{d}{dx}\left( uv \right)=uv'+u'v$ , in the right-hand side.
On differentiating with respect to x, we get:
$\dfrac{1}{z}\dfrac{dz}{dx}=x\times \dfrac{d}{dx}\left( \log x \right)+\log x$
$\Rightarrow \dfrac{1}{z}\dfrac{dz}{dx}=x\times \dfrac{1}{x}+\log x$
$\Rightarrow \dfrac{1}{z}\dfrac{dz}{dx}=1+\log x$
We substituted $z={{x}^{x}}$ .
$\Rightarrow \dfrac{d}{dx}{{x}^{x}}={{x}^{x}}\left( 1+\log x \right)$
Now , we will substitute the value of $\dfrac{d}{dx}({{x}^{x}})$ in equation $(1)$ .
On substituting the value of $\dfrac{d}{dx}({{x}^{x}})$ in equation $(1)$ , we get , $2{{x}^{x}}[1+\log x]{{x}^{x}}-\cot y2{{x}^{x}}(1+\log x)-2{{x}^{x}}(-\cos e{{c}^{2}}y)\dfrac{dy}{dx}=0$
$\Rightarrow 2{{x}^{2x}}(1+\log x)-\cot y2{{x}^{x}}(1+\log x)+2{{x}^{x}}\cos e{{c}^{2}}y\dfrac{dy}{dx}=0$
$\Rightarrow 2{{x}^{x}}({{x}^{x}}(1+\log x)-\cot y(1+\log x))+2{{x}^{x}}\cos e{{c}^{2}}y\dfrac{dy}{dx}=0$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{2{{x}^{x}}(1+\log x)(\cot y-{{x}^{x}})}{2{{x}^{x}}\cos e{{c}^{2}}y}$
Now , we will find the value of $y$ at $x=1$ .
We are given ${{x}^{2x}}-2{{x}^{x}}\cot y-1=0.........$ equation $(2)$ .
To find the value of $y$ at $x=1$ , we will substitute $x=1$ in equation $(2)$ .
On substituting $x=1$ in equation $(2)$ , we get ,
${{1}^{2.1}}-{{2.1}^{1}}\cot y-1=0$

& \Rightarrow -2\cot y=0 \\\ & \Rightarrow \cot y=0 \\\ & \Rightarrow y=\dfrac{\pi }{2} \\\ \end{aligned}$$ So , at $$x=1$$ , we can see the value of $$y=\dfrac{\pi }{2}$$ . Now , substituting $$x=1$$ and $$y=\dfrac{\pi }{2}$$ in the value of $$\dfrac{dy}{dx}$$ we get , $$\dfrac{dy}{dx}=\dfrac{2(1+0)(0-1)}{2\times 1}=\dfrac{-2}{2}=-1$$ . Hence , the value of the derivative of $${{x}^{2x}}-2{{x}^{x}}\cot y-1=0$$ at $$x=1$$ is equal to $$-1$$ . Hence, the correct option is option(d). Note: Substitutions should be done carefully, because any mistake during substitution will lead to wrong answers. Also, the value of the derivative of ${{x}^{x}}$ with respect to $x$ should be remembered as a formula. It is very helpful while solving difficult problems.